Normal contact force for a sliding slope

[Happiness]

For the following question, how do we know that the acceleration of $M$ is constant over time? And is the normal contact force between the two masses smaller as compared to that where $M$ is fixed?

The acceleration depends on the net force on $M$, which depends on the normal contact force $N$ between $m$ and $M$. The force $N$ depends on how tightly the two surfaces are pressed together. So it seems plausible that $N$ is smaller when $M$ is free to slide compared to that when $M$ is fixed.

The initial velocities of $m$ and $M$ are zero. Suppose we set up the initial conditions by holding $m$ and $M$ fixed and then release them. Would the initial (when both masses are just released from grip) value of $N$ in this case be the same as the value of $N$ where $M$ is always fixed?

I guess they would be the same. But since the acceleration is constant over time, the value of $N$, at all other time after $M$ starts to slide, must be the same as the initial value of $N$. Then, we must conclude that the value of $N$ where $M$ is fixed is equal to the value of $N$ where $M$ is sliding away from $m$. But when $M$ is sliding, it seems that the two masses are less tightly pressed together, and so $N$ should be smaller.

 

[wrobel]

you should write the equations of motion to this system.

 

[Happiness]

you should write the equations of motion to this system.

I get $N$ is constant provided $a_x$ or $A_x$ or $N'$ is constant, where $a$ and $A$ are the accelerations of $m$ and $M$, and $N'$ is the normal contact force on $M$ by the floor.

How do we know $a_x$ or $A_x$ or $N'$ is constant?

Suppose $N$ is velocity dependent. I believe the equations of motion, eg., $N\cos\theta+Mg=N'$, still holds, just that now $a_x$, $A_x$ and $N'$ will be time dependent. So it doesn't seem like we can deduce $N$ is time independent from the equations of motion.

 

[wrobel]

$$m\boldsymbol a_m=\boldsymbol N+m\boldsymbol g,\quad M\boldsymbol a_M=-\boldsymbol N+\boldsymbol R+M\boldsymbol g;\qquad (*)$$
$$\boldsymbol a_m=\boldsymbol a_M+w\boldsymbol e_x;\quad \boldsymbol N=N\boldsymbol e_y;$$
$$\boldsymbol n=\cos\theta\boldsymbol e_y-\sin\theta\boldsymbol e_x;\quad \boldsymbol e=\sin\theta \boldsymbol e_y+\cos\theta\boldsymbol e_x;$$
$$\boldsymbol a_M=a_M\boldsymbol e,\quad \boldsymbol R=R\boldsymbol n,\quad \boldsymbol g=-g\boldsymbol n$$
so we have four unknowns $w,a_M,R,N$ and four scalar equations (*)

 

[Happiness]

$$m\boldsymbol a_m=\boldsymbol N+m\boldsymbol g,\quad M\boldsymbol a_M=-\boldsymbol N+\boldsymbol R+M\boldsymbol g;\qquad (*)$$
$$\boldsymbol a_m=\boldsymbol a_M+w\boldsymbol e_x;\quad \boldsymbol N=N\boldsymbol e_y;$$
$$\boldsymbol n=\cos\theta\boldsymbol e_y-\sin\theta\boldsymbol e_x;\quad \boldsymbol e=\sin\theta \boldsymbol e_y+\cos\theta\boldsymbol e_x;$$
$$\boldsymbol a_M=a_M\boldsymbol e,\quad \boldsymbol R=R\boldsymbol n,\quad \boldsymbol g=-g\boldsymbol n$$
so we have four unknowns $w,a_M,R,N$ and four scalar equations (*)

There should be 5 unknowns because $m$ could accelerate vertically as it slides down the slope.

Anyway, my question is how do we know the terms in the equations of motion are time independent?

 

[wrobel]

mm could accelerate vertically

sure, that has already been taken into account, look at the formulas carefully

 

[Happiness]

sure, that has already been taken into account, look at the formulas carefully

Shouldn't it be $\mathbf {a_m}=\mathbf {a_M}+w\mathbf {e_x}+u\mathbf {e_y}$?

$M$ is free to slide too. It accelerates to the right.

 

[wrobel]

the term $w\boldsymbol e_x$ is the acceleration of $m$ relative to the wedge; $\boldsymbol a_M$ is acceleration of the wedge. We use the summation of accelerations theorem, the term $u\boldsymbol e_y$ is unnecessary

 

[Happiness]

the term $w\boldsymbol e_x$ is the acceleration of $m$ relative to the wedge; $\boldsymbol a_M$ is acceleration of the wedge. We use the summation of accelerations theorem, the term $u\boldsymbol e_y$ is unnecessary

What's the theorem about? The relative acceleration won't just be in the x direction, so I don't see why you only introduce a constant $w$.

Anyway, this doesn't answer the question of time independence.

 

[wrobel]

you do not understand banal things but what a self-confidence :)

 

[Happiness]

you do not understand banal things but what a self-confidence :)

You made a mistake isn't it? There should be 5 unknowns.

 

[wrobel]

this continues posting #4. Solving linear system (*) we get
$$w=cg(m+M)\sin\theta,\quad a_M=-\frac{c}{2}mg\sin 2\theta,\quad R=cMg(M+m),\quad N=c mMg\cos\theta, $$
here $$c=\frac{1}{M+m\sin^2\theta}.$$ Calculations are in the attachmentSimpler way to solve this problem is to use the Lagrange equations. There is another problem of the same type but little bit more complicated : to replace the block with a disk that rolls on the wedge without slipping.